I can do the first part of part b but don't know how to work out the other roots

And part c is just lost on me

Thanks

For the first part of (b), you noticed that is a root of the quartic by recognising that it gives .

But the same can be said for any , where is such that . Can you write down all such (between 0 and - why is this enough?), and consequently all the remaining roots of the quartic?

For part (c), write down a quartic with roots by making a substitution for into the quartic, rearranging and squaring suitably. Then recall that you can determine the sum of the roots of a polynomial from it's coefficients.

(Original post by Farhan.Hanif93)
For the first part of (b), you noticed that is a root of the quartic by recognising that it gives .

But the same can be said for any , where is such that . Can you write down all such (between 0 and - why is this enough?), and consequently all the remaining roots of the quartic?

For part (c), write down a quartic with roots by making a substitution for into the quartic, rearranging and squaring suitably. Then recall that you can determine the sum of the roots of a polynomial from it's coefficients.

I'm still stuck on part c. When you say write a quartic with roots by maing a substitution for into the quartic. Do you mean ? if so, where do I go from there.

(Original post by TheKian)
I'm still stuck on part c. When you say write a quartic with roots by maing a substitution for into the quartic. Do you mean ? if so, where do I go from there.

Not quite. Let t = √u. Then, after some rearrangement followed by careful squaring (i.e. to get rid of the square roots), you will have another quartic in terms of u with roots u = t^2 for each t that is a root of the old quartic.

I can do the first part of part b but don't know how to work out the other roots

And part c is just lost on me

Thanks

For part C it is just a disguised FP1 question where you find the relationship between the roots of the polynomial and the coefficients. If you remember that Σα2 = (Σαβ)2 -2Σαβ. Apply the same concept here and, and you will have to spot a symmetry between the roots you will get to the result.

(Original post by Farhan.Hanif93)
Not quite. Let t = √u. Then, after some rearrangement followed by careful squaring (i.e. to get rid of the square roots), you will have another quartic in terms of u with roots u = t^2 for each t that is a root of the old quartic.

That sounds like hard work. I think it is easier to let the roots be then form by considering the square of the sum of the roots - you can then use the various relationships between the coefficients and sum, product, etc of the roots.

(Original post by atsruser)
That sounds like hard work. I think it is easier to let the roots be then form by considering the square of the sum of the roots - you can then use the various relationships between the coefficients and sum, product, etc of the roots.

[Aside: latex still broken - FFS - someone please fix this]

It largely boils down to whether one is required to take the extra step in proving the identity relating the sum of the squares of the 4 roots to the sum and products of the roots at A-Level - I don't recall it as quotable when I did FP2 around 5 years ago but I may have misremembered.

If that is the case, I don't see how this is any easier than squaring both sides of u^2 - 6u + 1 = 4(1-u)√u and rearranging for a quartic that you may simply read the result off from; as opposed to reading two results from the old quartics, proving a relation (which requires squaring, too) and then computing the sum of the squares by substitution into said relation. [Although, I concede that this is probably the less obvious/unexpected approach.]

(Original post by Farhan.Hanif93)
It largely boils down to whether one is required to take the extra step in proving the identity relating the sum of the squares of the 4 roots to the sum and products of the roots at A-Level - I don't recall it as quotable when I did FP2 around 5 years ago but I may have misremembered.

If that is the case, I don't see how this is any easier than squaring both sides of u^2 - 6u + 1 = 4(1-u)√u and rearranging for a quartic that you may simply read the result off from; as opposed to reading two results from the old quartics, proving a relation (which requires squaring, too) and then computing the sum of the squares by substitution into said relation. [Although, I concede that this is probably the less obvious/unexpected approach.]